A secant line intersects the graph of $y=\log(x)$ at two points with $x$ -coordinates $8$ and $t$. What is the slope of the secant line? Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{\log(t)-\log(8)}{8}$ (Choice B) B $\dfrac{\log(t-8)}{t-8}$ (Choice C) C $\dfrac{\log(t-8)}{t}$ (Choice D) D $\dfrac{\log(t)-\log(8)}{t-8}$
We are given that the secant line intersects the graph at $x=8$ and $x=t$. Since these points are on the curve $y=\log(x)$, we know that their $y$ -values are $y=\log(8)$ and $y=\log(t)$, respectively. To summarize this part, we know that the secant line passes through the points $(8,\log(8))$ and $(t,\log(t))$. This should be enough to find the slope of that line. $\begin{aligned} \text{Slope}&=\dfrac{\text{Change in }y}{\text{Change in }x} \\\\ &=\dfrac{\log(t)-\log(8)}{t-8} \end{aligned}$ In conclusion, the slope of the secant line is $\dfrac{\log(t)-\log(8)}{t-8}$.